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Let $P\in\mathbb{R}^{n\times n}$ be a nontrivial idempotent matrix: $P^2=P$, $P\neq 0$, $P\neq I,$ where $I$ is the $n\times n$ identity matrix.

What are the eigenvalues of $P$?

Solution:
Let $x$ be an eigenvector of $P$, with corresponding eigenvalue $\lambda$: $$Px=\lambda x.$$ We also have $$P(Px)=P\lambda x = \lambda Px=\lambda^2x,$$ and, since $P^2=P,$ we have $$P^2x=Px\implies\lambda^2=\lambda.$$ So, $\lambda(\lambda-1)=0\implies \lambda=0$ or $\lambda=1.$

However, is it not also correct to say that, since $P$ is idempotent, we also have $$P^m=P^{m-1}P=P^{m-2}PP=P^{m-2}P=P^{m-3}P=...=P$$ for any integer $m\geq0$? If that is correct, then we also have $$P^mx=\lambda^mx,$$ and so $$\lambda^m=\lambda$$ or $\lambda(\lambda^{m-1}-1)=0\implies\lambda=0$ or $\lambda^{m-1}=1$. Is there anything wrong with saying $$\lambda=\exp\left(2\pi i\frac{k}{m-1}\right),$$ for $k=0,1,...,m-2$?

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    $\begingroup$ For $P$ such that $P^2 = P$, we of course have $P^k = P$ for $k \ge 2$, but we still have to contend with $P^2 = P$! $\endgroup$ Commented Feb 25, 2014 at 21:22
  • $\begingroup$ Yes, I guess that's the key. No matter what complex solutions we get for the case of $P^k=P$ so $\lambda^k=\lambda$, the only solutions that we can keep also have to satisfy $P^2=P,$ which requires $\lambda^2=\lambda$, which is only true for $\lambda=0$ or $\lambda=1$. Thanks! $\endgroup$
    – Mike Bell
    Commented Feb 26, 2014 at 9:35
  • $\begingroup$ I've been thinking of exactly the same problem as the OP and I don't understand the answers. Could someone explain? Please keep in mind that I'm a simple engineer $\endgroup$
    – NNN
    Commented Jun 27, 2021 at 3:24
  • $\begingroup$ I think I get it. I think it probably means that other eigenvalues apart from the pair (0,1) do not satisfy $(P-\lambda_1{I})(P-\lambda_2{I}) = 0$ $\endgroup$
    – NNN
    Commented Jun 27, 2021 at 4:20
  • $\begingroup$ Sorry for the repeated comments. But, after reflection, I don't understand the answers. Any help would be appreciated. What does polynomial annilates P mean? Or $\endgroup$
    – NNN
    Commented Jul 3, 2021 at 5:43

2 Answers 2

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Since we have $P^2=P$ then the polynomial $x^2-x=x(x-1)$ annihilates $P$ so the set of eigenvalues of $P$ is a subset of the set of the roots of this polynomial: $$\operatorname{sp}(P)\subset\{0,1\}$$

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  • $\begingroup$ Beat me to it *shakesfist* $\endgroup$
    – Emily
    Commented Feb 25, 2014 at 20:57
  • $\begingroup$ Thanks for the fast answer. What does it mean for a polynomial to annihilate P, and what is the meaning of $\mathrm{sp}(P)\subset\{0,1\}$? $\endgroup$
    – Mike Bell
    Commented Feb 25, 2014 at 21:05
  • $\begingroup$ @MikeBell It is the interpretation of the Cayley-Hamilton theorem over a field. $\endgroup$
    – Emily
    Commented Feb 25, 2014 at 21:07
  • $\begingroup$ Ok, now that I've thought about it more, I guess $\mathrm{sp}(P)\subset\{0,1\}$ is saying that $0,1\in$ the spectrum of $P$. But, as for the annihilation part, I guess I'll have to study the Cayley-Hamilton theorem. $\endgroup$
    – Mike Bell
    Commented Feb 26, 2014 at 9:38
  • $\begingroup$ You nailed it, my friend! $\endgroup$
    – amWhy
    Commented Feb 26, 2014 at 14:09
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Without appealing to other theorems or arguments, your analysis is fine until you say that $\lambda^{m-1} = 1$ therefore $\lambda = \exp 2\pi i \frac{k}{m-1}$.

The error here is that it excludes the possibility of repeated eigenvalues. That is, the set of all $\lambda \neq 0$ is not equal to the set (of unique values) generated by $\exp 2\pi i \frac{k}{m-1}$, but rather it must be a subset of the unique elements generated by $\exp 2\pi i \frac{k}{m-1}$. Since $P$ only has $n$ eigenvalues (including multiplicity), if one or more has multiplicity $> 1$, then there will be some values of $m,k$ such that $\exp 2\pi i \frac{k}{m-1}$ does not yield an eigenvalue.

So it would be more proper to say that $\lambda \subset \{\exp 2\pi i \frac{k}{m-1}\ \mid\ m,k \in \mathbb{Z}\},$ but this doesn't solve your problem.

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